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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{5}{x+1}+\dfrac{2x}{\left(x+1\right)\left(x-4\right)}=\dfrac{2}{x-4}\) ĐKXĐ \(\left\{{}\begin{matrix}x\ne-1\\x\ne4\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{5\left(x-4\right)}{\left(x+1\right)\left(x-4\right)}+\dfrac{2x}{\left(x+1\right)\left(x-4\right)}=\dfrac{2\left(x+1\right)}{\left(x-4\right)\left(x+1\right)}\)
`=> 5(x-4) +2x=2(x+1)`
`<=> 5x-20 +2x=2x+2`
`<=> 7x-20=2x+2`
`<=> 7x-2x=2+20`
`<=> 5x=22`
`<=>x=22/5`
Vậy pt đã cho có nghiệm `x=22/5`
![](https://rs.olm.vn/images/avt/0.png?1311)
=> (x+1)[5(x^2-x+1)-7x^2]=0
=>x+1=0 vì 5(x^2-x+1)-7x^2 khác 0
=> x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=x^2-1-\left(x^2+4x+4\right)=x^2-1-x^2-4x-4=11\)
\(\Leftrightarrow-5x=15\)
\(\Leftrightarrow x=-3\)
Vậy ...
\(b,=\left(x-3-2x+5\right)\left(x-3+2x-5\right)=0\)
\(\Leftrightarrow\left(-x+2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy ...
a) \(\left(x-1\right)\left(x+1\right)-\left(x+2\right)^2=11\)
\(\Rightarrow x^2-1-x^2-4x-4-11=0\)
=> -4x - 16 = 0
=> -4x = 16
=> x = -4
b) \(\left(x-3\right)^2-\left(2x-5\right)^2=0\)
=> (x - 3 + 2x - 5).(x - 3 - 2x + 5) = 0
=> (3x - 8).(-x + 2) = 0
=> x = 8/3 hoặc x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(\left(x+5\right)^3-x^3-125\)
\(=x^3+15x^2+75x+125-x^3-125\)
\(=15x^2+75x\)
\(5\left(x-5\right)-8\left(x-11\right)=0\)
\(\Leftrightarrow5x-25-8x+88=0\)
\(\Leftrightarrow\left(5x-8x\right)+\left(88-25\right)=0\)
\(\Leftrightarrow-3x+63=0\)
\(\Leftrightarrow-3x=0-63\)
\(\Leftrightarrow-3x=-63\)
\(\Leftrightarrow x=21\)
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